azota à không chỉ đâu

1.Điều kiện : \(x\ge0\)

\(\Rightarrow\hept{\begin{cases}x+3,4>0\\x+2,4>0\\x+7,2>0\end{cases}}\Rightarrow\hept{\begin{cases}\left|x+3,4\right|=x+3,4\\\left|x+2,4\right|=x+2,4\\\left|x+7,2\right|=x+7,2\end{cases}}\)

\(\Rightarrow\left|x+3,4\right|+\left|x+2,4\right|+\left|x+7,2\right|=x+3,4+x+2,4+x+7,2\)

                                                                                \(=3x+13=4x\)

\(\Rightarrow4x-3x=13\)

\(\Rightarrow x=13\)

Vậy \(x=13\)

2.\(3^{n+3}+3^{n+1}+2^{n+3}+2^{n+2}\)

\(=3^n\left(3^3+3\right)+2^n\left(2^3+2^2\right)\)

\(=3^n\left(27+3\right)+2^n\left(8+4\right)\)

\(=3^n.30+2^n.12\)

\(=6\left(3^n.5+2^n.2\right)⋮6\)

4.a)

• \(3^{34}=3^{30+4}=3^{30}.3^4=3^{3.10}.3^4=\left(3^3\right)^{10}.3^4=27^{10}.3^4\)

\(5^{20}=5^{2.10}=\left(5^2\right)^{10}=25^{10}\)

Vì \(27^{10}>25^{10}\Rightarrow27^{10}.3^4>25^{10}\)

hay \(3^{34}>5^{20}\)

• \(17^{20}=17^{4.5}=\left(17^4\right)^5=83521^5>71^5\)

b)\(2^{300}=2^{3.100}=\left(2^3\right)^{100}=8^{100}\)

\(3^{200}=3^{2.100}=\left(3^2\right)^{100}=9^{100}\)

Vì \(8^{100}< 9^{100}\Rightarrow2^{300}< 3^{200}\)

\(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\Rightarrow0,4< \left|x-\dfrac{7}{5}\right|< 0,6\)

\(\Rightarrow\left|x-\dfrac{7}{5}\right|=\dfrac{1}{2}\Rightarrow x-\dfrac{7}{5}=\left[{}\begin{matrix}\dfrac{1}{2}\\-\dfrac{1}{2}\end{matrix}\right.\)

\(\Rightarrow x=\left[{}\begin{matrix}\dfrac{1}{2}+\dfrac{7}{5}\\-\dfrac{1}{2}+\dfrac{7}{5}\end{matrix}\right.\Rightarrow x=\left[{}\begin{matrix}\dfrac{19}{10}\\\dfrac{9}{10}\end{matrix}\right.\)

Vậy \(x=\dfrac{19}{10}\) hoặc \(x=\dfrac{9}{10}\)

\(\dfrac{2}{5}< \left|x-\dfrac{7}{5}\right|< \dfrac{3}{5}\)

\(\Rightarrow0,4< \left|x-1,4\right|< 0,6\)

\(\Rightarrow\left|x-1,4\right|=0,5\)

\(\Rightarrow x-1,4=\left[{}\begin{matrix}0,5\\-0,5\end{matrix}\right.\)

\(\Rightarrow x=\left[{}\begin{matrix}0,5+1,4\\-0,5+1,4\end{matrix}\right.\)

\(\Rightarrow x=\left[{}\begin{matrix}1,9\\0,9\end{matrix}\right.\)

Vậy x = 1,9 hoặc x = 0,9

A=a=b=c=0 đó bạn ( mình ko bt cách giải)

Ta có:\(\left(-5a^2b^4c^6\right)^7-\left(9a^3bc^5\right)^8=0\)

\(\left(-5\right)^7a^{14}b^{28}c^{42}-9^8a^{24}b^8c^{40}=0\)

Vì \(a^{14}b^{28}c^{42}\ge0\Rightarrow\left(-5\right)^7a^{14}b^{28}c^{42}\le0\)

\(a^{24}b^8c^{40}\ge0\Rightarrow9^8a^{24}b^8c^{40}\ge0\)

\(\Rightarrow\left(-5\right)^7a^{14}b^{28}c^{42}-9^8a^{24}b^8c^{40}\le0\)

Mà VP=0

Dấu "=" xảy ra khi

\(\left(-5\right)^7a^{14}b^{28}c^{42}=0\) và \(9^8a^{24}b^8c^{40}=0\)

\(\Rightarrow a=b=c=0\)

\(\Rightarrow A=a+b+c=0+0+0=0\)

C, 1/8
d, 5